If $p^2\sinh x+q^2\cosh x = r^2$ has at least one root, how do I show that $r^4 > p^4-q^4?$
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Your equation is equivalent to:
$$(p^2+q^2)e^x+(p^2-q^2)e^{-x}=2r^2$$
or:
$$(p^2+q^2)e^{2x}-2r^2e^x+(p^2-q^2)=0\tag{1}$$
This is a quadratic equation with respect to $t=e^x$. If (1) has at least one root, it means that there has to be one positive solution of (1). And it is possible if:
$$D=(2r^2)^2-4(p^2+q^2)(p^2-q^2)\ge 0$$
...which leads directly to:
$$r^4\ge p^4-r^4$$
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